posted by Fai .
A definite amount of NH4cl was boiled with 100ml of 0.8N naoh for complete reaction. After the reaction, the reactant mixture containing excess of naoh was neutralized with 12.5ml of 0.75N nh2s04. Calculate the amount of nh4cl taken.
The answer 3.78g
# milliequivalents = 100 x 0.8 = 80 total
-# m.e. backtitrated with NaoH = 12.5 x 0.75 = 9.375
NH4Cl initially = 80-9.375= 70.625 then
70.625 x m.e.w NH4Cl (about 53.5) = abpit 3.8 g.